The forces between molecules of a fluid are called cohesive forces, the forces between the molecules of a fluid and a solid are called adhesive forces. The interaction of these forces results in surface tension and capillary, which are crucial for our discussion.
Through basic experiments of soap films on a wireframe, we can understand that surfaces exert force. We can further hypothesize that the liquid surface is in tension from this experiment, and we need to do work to increase the surface area.
A molecule in the midst of a liquid interacts with all its neighbor, i.e., it is attracted by the neighboring molecules equally in all the direction; However, a molecule on the surface has half of these interactions. Hence, the molecules on the surface are at higher energy states in comparison to internal molecules. The molecules of the liquid configure in such a way that the total energy is at minima. The surface energy of a liquid drop is proportional to number of surface molecules. Hence the liquid takes a configuration in which number of molecules on the surface are minimum, i.e., the surface area of the drop is minimum. To increase the surface area or to create new surface, additional energy will be required to bring the interior molecules on the surface. This energy will be equivalent to work required in breaking the bonds of interior molecules and bringing them on the surface where the bonds are half of the interior.
Suppose, we want to increase the surface area of a liquid by \(dA\). The work required to bring the additional molecules on the surface will be proportional to \(dA\) (the additional number of molecules), i.e., we can write:
\[\delta{W}=\gamma{dA}\]
where \(\gamma\) is the surface tension - the energy required to increase the surface area by one unit.
The surface tension is also defined as force per unit length. Consider the soap film on the wire frame attached with a moving rod (fig. 1). Initially, the rod is in equilibrium; the rod is perturbed by small displacement, \(dx\). The work done in this perturbation -
\[\delta{W}=Fdx-2\gamma{L}dx\]
Since the rod was in equilibrium -
\[\delta{W}=0\] \[F=2\gamma{L}dx\]
This suggests that the \(\gamma\) is the force exerted by the surface per unit length of the rod.
Typical values of surface tension tbl. 1 and a few remarks [1]:
Water has higher surface tension than other liquids (except for the liquid metals).
Other molecules often reduce the surface tension of water (except for some salts).
Increased temperature reduces the surface tension.
Electrical conditions near the surface can increase or reduce the surface tension.
| Liquid | Surface tension (dynes/cm) |
|---|---|
| Water | 73 |
| Salt water | 75 |
| Soapy water | 20-30 |
| Ether | 17 |
| Alcohol | 23 |
| Carbon tetrachloride | 27 |
| Lubricating oil | 25-35 |
| Mercury | 480 |
For a curved interface, the surface tension creates a pressure difference in fluids on both sides of the interface. In simple spherical drops, the pressure is higher on the concave side. This excess pressure has many consequences: formation of drops in a falling water jet, merging of smaller drops into larger ones, capillary adhesion (between two plates, between hairs, bewteen sand particles). In the following, we derive the pressure difference in terms of the surface tension.
From fig. 2, the work required to increase the volume of drop
\[\delta{W}=p_{o}\left(4{\pi}R^{2}\delta{R}\right)-p_{w}\left(4{\pi}R^{2}\delta{R}\right)-\gamma_{ow}\left(8{\pi}R\delta{R}\right)\]
At equilibrium, \(\delta{W}=0\).
\[\Delta{p}=p_{o}-p_{w}=\frac{2\gamma_{ow}}{R}\]
The above equation simply implies that for each spherical surface or interface, the pressure difference is \(\frac{2\gamma_{12}}{R}\).
A soap bubble is shown in fig. 3; there are two interfaces: inner and outer interface. For each interface, from inside to outside, the pressure increases by \(\frac{2\gamma_{sa}}{R}\); hence
\[\Delta{p}=p_{o}-p_{i}=\frac{4\gamma_{sa}}{R}\]
A small portion of a generic surface is shown in the fig. 4 with primary radii of curvature as \(R_{1}\) and \(R_{2}\). In the following, we assume that the shape of surface is preserved for an arbitrarily small perturbation in which radius of curvature is increased by a small amount \(\delta{z}\).
To preserve the shape, we need to ensure (from the fig. 4)
\[\frac{\delta{x_{1}}}{R_{1}}=\frac{\delta{x_{1}}+\Delta{\left(\delta{x_{1}}\right)}}{R_{1}+\delta{z}}\]
similarly,
\[\frac{\delta{x_{2}}}{R_{2}}=\frac{\delta{x_{2}}+\Delta{\left(\delta{x_{2}}\right)}}{R_{2}+\delta{z}}\]
The initial area is -
\[dA=\delta{x_{1}}\delta{x_{2}}\]
Area after the perturbation -
\[d\tilde{A}=\left(\delta{x_{1}}+\Delta{\left(\delta{x_{1}}\right)}\right)\left(\delta{x_{2}}+\Delta{\left(\delta{x_{2}}\right)}\right)\]
The increase in the area -
\[\delta{dA}=\left(1+\frac{\delta{z}}{R_{1}}\right)\left(1+\frac{\delta{z}}{R_{2}}\right)\delta{x_{1}}\delta{x_{2}}-\delta{x_{1}}\delta{x_{2}}\]
The work require to perturb the surface -
\[\delta{W}=-p_{i}\delta{x_{1}}\delta{x_{2}}\delta{z}+p_{o}\delta{x_{1}}\delta{x_{2}}\delta{z}+\gamma\delta{dA}\]
At equilibrium -
\[\delta{W}=0\]
\[\left(p_{i}-p_{o}\right)\delta{z}=\gamma\left[\frac{\delta{z}}{R_{1}}+\frac{\delta{z}}{R_{2}}+\frac{\left(\delta{z}\right)^{2}}{R_{1}R_{2}}\right]\]
Since \(\delta{z}\ll\ R_{1}R_{2}\),
\[\left(p_{i}-p_{o}\right)=\gamma\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\]
fig. 5 shows a drop of liquid mashed between two plates. The drop will take a form of funicular cylinder of radius \(R\) and height \(H\). If we look from inside of the drop; there are two curvatures: concave one with radius R and convex one with radius \(\frac{H}{2\cos{\theta_{E}}}\). Hence pressure difference between the liquid in the drop and outside gaseous medium is -
\[\Delta{p}=\gamma\left(\frac{1}{R}-\frac{1}{\frac{H}{2\cos{\theta_{E}}}}\right)\]
As \(R\) is sufficiently large -
\[\Delta{p}=-\frac{2\gamma{\cos{\theta_{E}}}}{H}\]
The negative pressure acts as suction and attracts the plate towards each other by the force -
\[F=\pi{R^{2}}\frac{2\gamma{\cos{\theta_{E}}}}{H}\]
Note - Rememeber that the Pascal law is only true when the surface is in equilibrium.
A cylindrical jet of water of radius \(r\) tend to break apart into drops; this can be explained using the Laplace law. As the cyliderical jet of water has radius \(r\); the pressure inside the drop is \(\frac{\gamma}{r}\) in access to atmospheric pressure. As there are no perfectly cylindrical jets, there are often irregularities which cause significantly high pressure and result in breakage of jet into drops. Further, the surface area of resulting drops should be less than the original surface area of cylindrical jet: the droplets form to minimze the surface area. Let us consider, the original jet of radius \(R\) and length \(L\) which resuls in \(n\) droplets of radius \(r\). The ratio of original surface area to droplets surface area -
\[\frac{S_{0}}{S_{n}}=\frac{2\pi{R}L}{4\pi{n}r^{2}}\]
From the conservation of volume (assuming density does not change) -
\[\pi{R^{2}}L=\frac{4}{3}\pi{n}r^{3}\]
This results in -
\[\frac{S_{0}}{S_{n}}=\frac{2r}{3R}\]
Since \(S_{0}>{S_{n}}\) -
\[r>\frac{3R}{2}\]
This suggests that the radius of droplets will be higher than the original jet.
Consider a drop of liquid lying on a fibre pipe of radius \(b\); the contact angle at the interface is zero, i.e., the liquid is able to wet the fibre. From the Pascal law at a point on the surface of the drop (refer fig. 6) -
\[\Delta{p}=\gamma\left(\frac{1}{\frac{z}{\cos{\theta}}}-\frac{\ddot{z}}{\sqrt{\left(1+\dot{z}^{2}\right)^{3}}}\right)\]
The equation of equilibrium at each point of the surface of drop is -
\[\left(\frac{1}{z\sqrt{1+\dot{z}^{2}}}-\frac{\ddot{z}}{\sqrt{\left(1+\dot{z}^{2}\right)^{3}}}\right)=\frac{\Delta{p}}{\gamma}\]
Further, we can derive a first order ordinary differential equation representing the surface of the drop by considering the equilibrium of a part of the drop, as shown in fig. 6.
\[-2\pi\gamma{b}+2\pi\gamma{z}\cos{\theta}-\pi\left(z^{2}-b^{2}\right)\Delta{p}=0\]
\[z\cos{\theta}-b-\left(z^{2}-b^{2}\right)\frac{\Delta{p}}{2\gamma}=0\]
\[\frac{z}{\sqrt{1+\dot{z}^{2}}}-b-\left(z^{2}-b^{2}\right)\frac{\Delta{p}}{2\gamma}=0\]
At \(z=L\), \(\dot{z}=0\) -
\[L-b-\left(L^{2}-b^{2}\right)\frac{\Delta{p}}{2\gamma}=0\]
The excess pressure -
\[\Delta{p}=2\gamma\frac{1}{L+b}\]
Wetting is the phenomena of spreading of liquid on a solid substrate. Wetting can be characterized into -
Total wetting - liquid has strong affinity to solid substrate.
Partial wetting - otherwise.
These caan be identified using the spreading parameter.
The spreading parameter is difference in the surface energy per unit area of interface before and after the deposition of the drop.
\[S=\gamma_{SG}-\left(\gamma_{SL}+\gamma_{LG}\right)\]
Total wetting occurs when the surface energy decreases, i.e., \(S>0\); else partial wetting happens with \(S<0\). In total wetting, the drop completely spreads over the surface in other to minimize the surface energy; whereas, in partial wetting, the drop forms a curved cap at equilibrium resulting in a contact angle \(\theta_{E}\). We can further say -
Using the force balance at line of contact or triple line, we can obtain a relation for contact angle \(\theta_{E}\) as
\[\gamma_{SG}-\left(\gamma_{SL}+\gamma_{LG}\cos{\theta_{E}}\right)=0\]
\[\gamma_{LG}\cos{\theta_{E}}=\gamma_{SG}-\gamma_{SL}\]
We can further write -
\[S=\gamma_{LG}\left(\cos{\theta_{E}}-1\right)\]
or
\[\cos{\theta_{E}}=1+\frac{S}{\gamma_{LG}}\]
Clearly, \(\theta_{E}\) is not defined when \(S>0\).
We can also derive this relation by using stationarity condition for \(W\) at equilibrium.
\[\delta{W}=\left(\gamma_{SG}-\gamma_{SL}\right)dx-\gamma_{LG}dx\cos{\theta_{E}}\]
At equilibrium -
\[\delta{W}=0\]
The capillary action or capillarity is the phenomena of a liquid flowing into narrow cavities or channels against the direction of gravity due to interfacial tension. Read the excerpt from [2] about the initial investigation on capoillarity -
with a view to explain arise apparent anomalies. After long groping in the dark, it was found to be desirable to discover by experiment what were the actual phenomena which required explanation. Hawksbee found that the height to which a fluid would rise in From a capillary tube of given radius was the same for all . thicknesses of the tube. was apparent that the attracting force of the tube was situated at or near the inner surface of the tube. But he does not appear to have taken account of the mutual attractions of the particles of the fluid. Jurin also found that the height of the column of fluid supported by capillary action depended solely upon the interior diameter of the tube at the upper surface of the fluid. From this he con- cluded that the column of fluid raised’ by Capillary Action was supported by the attraction of the periphery or section of the tube to which the upper surface of the fluid cohered or was contiguous.
Clairaut was the first to attempt to explain capillary phenomena on right principles, by referring them to the mutual attraction of the particles of the fluid, and to the attraction of the particles of the solid on the particles of the fluid; and supposing these attractions to depend upon the same function of the distance, he concludes that even if the attraction of the capillary tube be of a less intensity than that of the water, provided the intensity of the latter attraction be not twice as great as that of the former, the water will still rise in the tube (p. 121). Clairaut supposed that the attraction was sensible only at very small distances (p. 113).
Laplace (1706) realized that the liquid rise in a capillary tube if the dry surface energy of the tube, \(\gamma_{SG}\), is greater than wet surface energy, \(\gamma_{SL}\), per unit area [3]. Similar to the spreading parameter, an imbibition parameter, I, is defined as
\[I=\gamma_{SG}-\gamma_{SL}\]
\[I=\gamma_{LG}\cos{\theta}\]
when I is positive \(\rightarrow\) liquid rises, otherwise it drops in the tube.
Near a miniscus, the capillary force dominates till the capillary length after which gravity dominates. Capillary length can be estimated by -
\[\rho{g}\kappa^{-1}=\frac{\gamma}{\kappa^{-1}}\]
\[\kappa^{-1}=\sqrt{\frac{\gamma}{\rho{g}}}\]
From the fig. 1, we can derive the following -
\[p_{A}-p_{atm}=-\gamma\frac{\ddot{z}}{\sqrt{\left(1+\dot{z}^{2}\right)^{3}}}\]
At the capillary length, \(\dot{z}\ll{1}\) -
\[\gamma\ddot{z}-\rho{g}z=0\]
\[z=z_{0}\exp{\left(-\kappa{x}\right)}\]
We can find the capillary rise using the Laplace pressure law. The pressure near the meniscus on the liquid side of the interface is -
\[p=p_{atm}-\frac{2\gamma\cos{\theta_{E}}}{R}\]
also
\[p=p_{atm}-\rho{g}H\]
From above equations -
\[H=\frac{2\gamma{R}\cos{\theta_{E}}}{\rho{g}}\]
This can also be derived using equilinrium of forces and by minimization of energy.
fig. 11 shows a capillary bridge between two spherical particles. Understanding of capillary between two particles is important in understanding the flow behaviour in unsaturated medium, tensile strength, and moisture retention of unsaturated soils. The shape of capillary bridge is unknown apriori and is a function of contact angle, separation between the spheres (representing the roughness), surface tension, and volume of the liquid. This problem can be solved using the minimization of surface area with the given volume and further constraints on shape of sphere and contact angle. Here we will attempt the solution using the equilibrium method (we could also solve it using the Laplace law).
Consider a section ABCD of the capillary bridge shown in fig. 5. As the capillary bridge in equilibrium any part of it will also be in equilibrium -
\[-\left[2\pi{y_{0}}\right]\gamma+\left[2\pi{y}\right]\gamma\cos{\alpha}+\left[\pi{y^{2}}-\pi{y_{0}}^{2}\right]\psi=0\]
Substituting \(\cos{\alpha}=\frac{1}{\sqrt{1+\dot{y}^{2}}}\) and simplifying further, we get -
\[\frac{dy}{dx}=\sqrt{\frac{y^{2}}{\left(y_{0}+\frac{\psi}{2\gamma}\left(y_{0}^{2}-y^{2}\right)\right)^{2}}-1}\]
We can solve the above ordinary differential equation numerically using the boundary condition [4]:
The second boundary condition can be used to determine the capillary sunction \(\psi\).
Density of the soil particles is given as
\[\rho_{s}=\frac{M_{s}}{V_{s}}\]
Density of the air phase is given by
\[\rho_{a}=\frac{M_{a}}{V_{a}}\]
The specific gravity is defined as the ration of the density of the solid and density of air at \(4^\circ\)C and at 1 atm pressure
\[G_{s}=\frac{\rho_{s}}{\rho_{w}}\]
Air behaves as mixture of several gases and also varying amounts of water vapour.
Air is found in two states - Dry air (without water vapour) - Moist air (with water vapour)
Air behaves like an ideal gas and the behaviour is governed by
\[\bar{u}_{a}V_{a}=\frac{M_{a}}{w_{a}}RT\]
where, \(\bar{u}_{a}\) is the absolute air pressure, \(V_{a}\) is the volume of the air, \(M_{a}\) is the mass of the air, \(w_{a}\) is the molecular mass of the air, \(R\) is the universal gas constant, \(T\) is the absolute temperature.
In an isolated system, air follows the Boyle’s law
\[\bar{u}_{a_{1}}V_{a_{1}}=\bar{u}_{a_{2}}V_{a_{2}}\]
Density of the air is given by
\[\rho_{a}=\frac{M_{a}}{V_{a}}=\frac{\bar{u}_{a}w_{a}}{RT}\]
The concentration of the water vapour in the air is commonly expressed in terms of the relative humidity
\[RH=\frac{100\bar{u}_{v}}{\bar{u}_{v0}}\]
where, \(\bar{u}_{a}\) partial pressure of water vapour at the same temperature, \(\bar{u}_{v0}\) is the saturation pressure of water vapour at the same temperature.
The density of air decreases as the relative humidity increases i.e. moist air is lighter than the dry air.